GaloisTheory - Singaporefrederique/chap4.pdf · Chapter 7 GaloisTheory Galois theory is named after...

Chapter 7Galois Theory

Galois theory is named after the French mathematician Evariste Galois.Galois was born in 1811, and had what could be called the life of a misun-

derstood genius. At the age of 15, he was already reading material written forprofessional mathematicians. He took the examination to the “Ecole Polytech-nique” to study mathematics but failed and entered the “Ecole Normale” in1828. He wrote his first paper at the age of 18. He tried to advertise his work,and sent his discoveries in the theory of polynomial equations to the Academyof Sciences, where Cauchy rejected his memoir. He did not get discouraged, andin 1830, he wrote again his researches to the Academy of Sciences, where thistime Fourier got his manuscript. However, Fourier died before reading it.

A year later, he made a third attempt, and sent to the Academy of Sciencesa memoir called “On the conditions of solvability of equations by radicals”.Poisson was a referee, and he answered several months later, declaring the paperincomprehensible.

In 1832, he got involved in a love affair, but got rejected due to a rival, whochallenged him to a duel. The night before the duel, he wrote a letter to hisfriend, where he reported his mathematical discoveries. He died during the duelwith pistols in 1832.

It is after his death that his friend insisted to have his letter published, whichwas finally done by the mathematician Chevalier.

7.1 Galois group and fixed fields

Definition 7.1. If E/F is normal and separable, it is said to be a Galoisextension, or alternatively, we say that E is Galois over F .

Take E/F a Galois extension of degree n. Since it is separable of degree n, weknow that there are exactly n F -monomorphisms of E into an algebraic closureC. But E/F being also normal, every F -automorphism into C is actually and

173

174 CHAPTER 7. GALOIS THEORY

Figure 7.1: Evariste Galois (1811-1832)

F -automorphism of E. Thus there are exactly n = [E : F ] F -automorphisms ofE.

We can define the notion of a Galois group for an arbitrary field extension.

Definition 7.2. If E/F is a field extension, the Galois group of E/F , denotedby Gal(E/F ), is the set of F -automorphisms of E. It forms a group under thecomposition of functions.

Example 7.1. If E = Q( 3√2), then Gal(E/Q) = {1}, that is the identity on E.

The above example illustrates the fact that though one can always define aGalois group, we need the extension to be actually Galois to say that the orderof the Galois group is actually the degree of the field extension.

Definition 7.3. Let G = Gal(E/F ) be the Galois group of the extension E/F .If H is a subgroup of G, the fixed field of H is the set of elements fixed by everyautomorphism in H, that is

F(H) = {x ∈ E, σ(x) = x for all σ ∈ H}.

Vice-versa, if K is an intermediate field, define

G(K) = Gal(E/K) = {σ ∈ G, σ(x) = x for all x ∈ K}.

It is the group fixing K.

Galois theory has much to do with studying the relations between fixed fieldsand fixing groups.

Proposition 7.1. Let E/F be a finite Galois extension with Galois group G =Gal(E/F ). Then

1. The fixed field of G is F .

7.1. GALOIS GROUP AND FIXED FIELDS 175

2. If H is a proper subgroup of G, then the fixed field F(H) of H properlycontains F .

Proof. 1. Let F0 be the fixed field of G (and we have the field extensionsE/F0/F ). We want to prove that F0 = F .

We first note that if σ is an F -automorphism of E (that is σ is in G),then by definition of F0, σ fixes everything in F0, meaning that σ is anF0-automorphism. Thus the F -automorphisms in the group G coincidewith the F0-automorphisms in the group G.

Now we further have that E/F0 is Galois: indeed, we have E/F0/F withE/F Galois thus normal and separable, and E/F0 inherits both properties.

We now look at the degrees of the extensions considered:

|Gal(E/F0)| = [E : F0], |Gal(E/F )| = [E : F ],

since both are Galois. Furthermore by the first remark, the number of F−and F0− automorphisms in G coincide:

|Gal(E/F0)| = |Gal(E/F )|

showing that[E : F0] = [E : F ]

and by multiplicativity of the degrees

[E : F ] = [E : F0][F0 : F ] ⇒ [F0 : F ] = 1

and F = F0.

2. In order to prove that F ( F(H), let us assume by contradiction thatF = F(H).

Since we consider a finite Galois extension, we can invoke the Theorem ofthe Primitive Element and claim that

E = F (α), α ∈ E. (7.1)

Consider the polynomial

f(X) =∏

σ∈H

(X − σ(α)) ∈ E[X].

It is a priori in E[X], but we will prove now that it is actually in F [X].Since by contradiction we are assuming that F = F(H), it is enough toproof that f(X) is fixed by H. Indeed, take τ ∈ H, then

∏

σ∈H

(X − τσ(α)) =∏

σ∈H

(X − σ(α))


since τσ ranges over all H as does σ.

Thus f(X) ∈ F [X] and f(α) = 0 (σ must be the identity once whileranging through H). Now on the one hand, we have

deg f = |H| < |G| = [E : F ]

since we assume that H is proper and E/F is Galois. On the other hand,

deg f ≥ [F (α) : F ] = [E : F ]

since f is a multiple of the minimal polynomial of α over F (equality holdsif f is the minimal polynomial of α over F ), and E = F (α) by (7.1). Wecannot possibly have deg f < [E : F ] and deg f ≥ [E : F ] at the sametime, which is a contradiction and concludes the proof.

7.2 The fundamental Theorem of Galois theory

The most significant discovery of Galois is that (surely not in these terms!)under some hypotheses, there is a one-to-one correspondence between

1. subgroups of the Galois group Gal(E/F )

2. subfields M of E such that F ⊆M .

The correspondence goes as follows:

• To each intermediate subfield M , associate the group Gal(E/M) of allM -automorphisms of E:

G = Gal : {intermediate fields} → {subgroups of Gal(E/F )}M 7→ G(M) = Gal(E/M).

• To each subgroup H of Gal(E/F ), associate the fixed subfield F(H):

F : {subgroups of Gal(E/F )} → {intermediate fields}H 7→ F(H).

We will prove below that, under the right hypotheses, we actually have abijection (namely G is the inverse of F). Let us start with an example.

Example 7.2. Consider the field extension E = Q(i,√5)/Q. It has four Q-

automorphisms, given by (it is enough to describe their actions on i and√5):

σ1 : i 7→ i,√5 7→

√5

σ2 : i 7→ −i,√5 7→

√5

σ3 : i 7→ i,√5 7→ −

√5

σ4 : i 7→ −i,√5 7→ −

√5

7.2. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 177

thusGal(E/Q) = {σ1, σ2, σ3, σ4}.

The proper subgroups of Gal(E/Q) are

{σ1}, {σ1, σ2}, {σ1, σ3}, {σ1, σ4}

and their corresponding subfields are

E, Q(√5), Q(i), Q(i

√5).

We thus get the following diagram:

E

Q(√5) Q(i) Q(i

√5)

Q

<σ3>@@@

<σ4>��

<σ2>

@@@2

2�

�� 2

Theorem 7.2. Let E/F be a finite Galois extension with Galois group G.

1. The map F is a bijection from subgroups to intermediate fields, with in-verse G.

2. Consider the intermediate field K = F(H) which is fixed by H, and σ ∈ G.Then the intermediate field

σK = {σ(x), x ∈ K}

is fixed by σHσ−1, namely σK = F(σHσ−1).

Proof. 1. We first consider the composition of maps

H → F(H) → GF(H).

We need to prove that GF(H) = H. Take σ in H, then σ fixes F(H) bydefinition and σ ∈ Gal(E/F(H)) = G(F(H)), showing that

H ⊆ GF(H).

To prove equality, we need to rule out the strict inclusion. If H werea proper subgroup of G(F(H)), by the above proposition the fixed fieldF(H) of H should properly contain the fixed field of GF(H) which isF(H) itself, a contradiction, showing that

H = GF(H).


Now consider the reverse composition of maps

K → G(K) → FG(K).

This time we need to prove that K = FG(K). But

FG(K) = fixed field by Gal(E/K)

which is exactly K by the above proposition (its first point).

2. It is enough to compute F(σHσ−1) and show that it is actually equal toσK = σF(H).

F(σHσ−1) = {x ∈ E, στσ−1(x) = x for all τ ∈ H}= {x ∈ E, τσ−1(x) = σ−1(x) for all τ ∈ H}= {x ∈ E, σ−1(x) ∈ F(H)}= {x ∈ E, x ∈ σ(F(H))} = σ(F(H)).

We now look at subextensions of the finite Galois extension E/F and askabout their respective Galois group.

Theorem 7.3. Let E/F be a finite Galois extension with Galois group G. LetK be an intermediate subfield, fixed by the subgroup H.

1. The extension E/K is Galois.

2. The extension K/F is normal if and only if H is a normal subgroup of G.

3. If H is a normal subgroup of G, then

Gal(K/F ) ≃ G/H = Gal(E/F )/Gal(E/K).

4. Whether K/F is normal or not, we have

[K : F ] = [G : H].

Proof. 1. That E/K is Galois is immediate from the fact that a subextensionE/K/F inherits normality and separability from E/F .

2. First note that σ is an F -monomorphism of K into E if and only if σ isthe restriction to K of an element of G: if σ is an F -monomorphism of Kinto E, it can be extended to an F -monomorphism of E into itself thanksto the normality of E. Conversely, if τ is an F -automorphism of E, thenσ = τ |K is surely a F -monomorphism of K into E.

Now, this time by a characterization of a normal extension, we have

K/F normal ⇐⇒ σ(K) = K for all σ ∈ G.

7.3. FINITE FIELDS 179

Since K = F(H), we just rewrite

K/F normal ⇐⇒ σ(F(H)) = F(H) for all σ ∈ G.

Now by the above theorem, we know that σ(F(H)) = F(σHσ−1), and wehave

K/F normal ⇐⇒ F(σHσ−1) = F(H) for all σ ∈ G.

We are almost there, we now use again the above theorem that tells usthat F is invertible, with inverse G, to get the conclusion:

K/F normal ⇐⇒ σHσ−1 = H for all σ ∈ G.

3. To prove this isomorphism, we will use the 1st isomorphism Theorem forgroups. Consider the group homomorphism

Gal(E/F ) → Gal(K/F ), σ 7→ σ|K .

This map is surjective (we showed it above, when we mentioned that wecan extend σ|K to σ. Its kernel is given by

Ker = {σ, σ|K = 1} = H = Gal(E/K).

Applying the 1st isomorphism Theorem for groups, we get

Gal(K/F ) ≃ Gal(E/F )/Gal(E/K).

4. Finally, by multiplicativity of the degrees:

[E : F ] = [E : K][K : F ].

Since E/F and E/K are Galois, we can rewrite

|G| = |H|[K : F ].

We conclude by Lagrange Theorem:

[G : H] = |G|/|H| = [K : F ].

7.3 Finite fields

We will provide a precise classification of finite fields.

Theorem 7.4. Let E be a finite field of characteristic p.

1. The cardinality of E is|E| = pn,

for some n ≥ 1. It is denoted E = Fpn .


2. Furthermore, E is the splitting field for the separable polynomial

f(X) = Xpn −X

over Fp, so that any finite field with pn elements is isomorphic to E. Infact, E coincides with the set of roots of f .

Proof. 1. Let Fp be the finite field with p elements, given by the integersmodulo p. Since E has characteristic p, it contains a copy of Fp. Thus Eis a field extension of Fp, and we may see E as a vector space over Fp. Ifthe dimension is n, then let α1, . . . , αn be a basis. Every x in E can bewritten as

x = x1α1 + · · ·+ xnαn

and there are p choices for each xi, thus a total of pn different elementsin E.

2. Let E× be the multiplicative group of non-zero elements of E. If α ∈ E×,then

αpn−1 = 1

by Lagrange’s Theorem, so that

αpn

= α

for all α in E (including α = 0). Thus each element of E is a root of f ,and f is separable.

Now f has at most pn distinct roots, and we have already identified thepn elements of E as roots of f .

Corollary 7.5. If E is a finite field of characteristic p, then E/Fp is a Galoisextension, with cyclic Galois group, generated by the Frobenius automorphism

σ : x 7→ σ(x) = xp, x ∈ E.

Proof. By the above proposition, we know that E is a splitting field for a sepa-rable polynomial over Fp, thus E/Fp is Galois.

Since xp = x for all x in Fp, we have that

Fp ⊂ F(〈σ〉)

that is Fp is contained in the fixed field of the cyclic subgroup generated by theFrobenius automorphism σ. But conversely, each element fixed by σ is a rootof Xp −X so F(〈σ〉) has at most p elements. Consequently

Fp = F(〈σ〉)

andGal(E/Fp) = 〈σ〉.


This can be generalized when the base field is larger than Fp.

Corollary 7.6. Let E/F be a finite field extension with |E| = pn and |F | = pm.Then E/F is a Galois extension and m|n. Furthermore, the Galois group iscyclic, generated by the automorphism

τ : x 7→ τ(x) = xpm

, x ∈ E.

Proof. If the degree [E : F ] = d, then every x in E can be written as

x = x1α1 + · · ·+ xdαd

and there are pm choices for each xi, thus a total of

(pm)d = pn

different elements in E, so that

d = m/n and m|n.

The same proof as for the above corollary holds for the rest.

Thus a way to construct a finite field E is, given p and n, to constructE = Fpn as a splitting field for Xpn −X over Fp.

Theorem 7.7. If G is a finite subgroup of the multiplicative group of an arbi-trary field, then G is cyclic. Thus in particular, the multiplicative group E× ofa finite field E is cyclic.

Proof. The proof relies on the following fact: if G is a finite abelian group, itcontains an element g whose order r is the exponent of G, that is, the leastcommon multiple of the orders of all elements of G.

Assuming this fact, we proceed as follows: if x ∈ G, then its order divides rand thus

xr = 1.

Therefore each element of G is a root of Xr − 1 and

|G| ≤ r.

Conversely, |G| is a multiple of the order of every element, so |G| is at least asbig as their least common multiple, that is

|G| ≥ r

and|G| = r.

Since the order of |G| is r, and it coincides with the order of the element gwhose order is the exponent, we have that G is generated by g, that is G = 〈g〉is cyclic.


Since E× is cyclic, it is generated by a single element, say α:

E = Fp(α)

and α is called a primitive element of E. The minimal polynomial of α is calleda primitive polynomial.

Example 7.3. Consider the following irreducible polynomial

g(X) = X4 +X + 1

over F2. Let α be a root of g(X). A direct computation shows that α isprimitive:

α0 = 1, . . . , α4 = α+ 1, . . . , α7 = α3 + α+ 1, . . . , α14 = 1 + α3.

7.4 Cyclotomic fields

Definition 7.4. A cyclotomic extension of a field F is a splitting field E forthe polynomial

f(X) = Xn − 1

over F . The roots of f are called nth roots of unity.

The nth roots of unity form a multiplicative subgroup of the group E× ofnon-zero elements of E, and thus must be cyclic. A primitive nth root of unityis an nth root of unity whose order in E× is n. It is denoted ζn.

From now on, we will assume that we work in a characteristic char(F ) suchthat char(F ) does not divide n. (Otherwise, we have n = mchar(F ) and 0 =ζnn − 1 = (ζm − 1)char(F ) and the order of ζn is less than n.)

Example 7.4. The field Q(ζp) where p is a prime and ζp is a primitive pth rootof unity is a cyclotomic field over Q.

Let us look at the Galois group Gal(E/F ) of the cyclotomic extension E/F .Then σ ∈ Gal(E/F ) must map a primitive nth root of unity ζn to anotherprimitive nth root of unity ζrn, with (r, n) = 1. We can then identify σ with r,and this shows that

Gal(E/F ) ≃ Un

where Un denotes the group of units modulo n. This shows that the Galoisgroup is abelian.

Example 7.5. Consider the field extension Q(ζ3)/Q. We have

X3 − 1 = (X − 1)(X2 +X + 1).

The Galois group is given by:

σ : ζ3 7→ ζ23

σ2 : ζ3 7→ ζ3

7.4. CYCLOTOMIC FIELDS 183

and the group U3 of units modulo 3 is U3 = {1, 2}. Thus

Gal(Q(ζ3)/Q) = {σ, 1} ≃ {2, 1} = (Z/3Z)×.

Finally, since E/F is Galois (under the above assumption)

[E : F ] = |Gal(E/F )| = ϕ(n)

where ϕ(n) is the Euler totient function.From now on, we fix the base field F = Q. This means that a primitive nth

root of unity ζn is given by

ζn = ei2πr/n, (r, n) = 1.

Definition 7.5. The nth cyclotomic polynomial is defined by

Ψn(X) =∏

(i,n)=1

(X − ζin),

where the product is taken over all primitive nth roots of unity in C.

The degree of Ψn(X) is thus

deg(Ψn) = ϕ(n).

Example 7.6. For n = 1, 2, we have

Ψ1(X) = X − 1, Ψ2(X) = X − (−1) = X + 1.

Computing a cyclotomic polynomial is not that easy. Here is a formula thatcan help.

Proposition 7.8. We have

Xn − 1 =∏

d|nΨd(X).

In particular, if n = p a prime, then d is either 1 or p and

Xp − 1 = Ψ1(X)Ψp(X) = (X − 1)Ψp(X)

from which we get

Ψp(X) =Xp − 1

X − 1= Xp−1 +Xp−2 + · · ·+X + 1.

Proof. We prove equality by comparing the roots of both monic polynomials.If ζ is a nth root of unity, then by definition

ζnn = 0

and its order d divides n. Thus ζ is actually a primitive dth root of unity, anda root of Ψd(X).

Conversely, if d|n, then any root of Ψd(X) is a dth root hence a nth root ofunity.


Examples 7.7. For n = 3 and 5, we have a prime and thus we can use theabove formula:

Ψ3(X) = X2 +X + 1

Ψ5(X) = X4 +X3 +X2 +X + 1.

For n = 4 the primitive 4rth roots of unity are ±i, and by definition

Ψ4(X) = (X − i)(X + i) = X2 + 1.

Finally for n = 6, the possible values for d are 1,2,3 and 6. Thus

Ψ6(X) =X6 − 1

(X − 1)(X + 1)(X2 +X + 1)= X2 −X + 1.

From the above examples, it is tempting to say that in general Ψn(X) hasinteger coefficients. It happens to be true.

Proposition 7.9. The nth cyclotomic polynomial Ψn(X) satisfies

Ψn(X) ∈ Z[X].

Proof. We proceed by induction on n. It is true for n = 1 since X − 1 ∈ Z[X].Let us suppose it is true for Ψk(X) where k is up to n− 1, and prove it is alsotrue for n.

Using the above proposition, we know that

Xn − 1 =∏

d|nΨd(X)

= Ψn(X)∏

d|n,d<n

Ψd(X).

The aim is to prove that Ψn(X) ∈ Z[X]:

Ψn(X) =Xn − 1

∏

d|n,d<n Ψd(X).

First note that Ψn(X) has to be monic (by definition), and both Xn − 1 andΨd(X) (by induction hypothesis) are in Z[X]. We can thus conclude invokingthe division algorithm for polynomials in Z[X].

We conclude by proving the irreducibility of the cyclotomic polynomials.

Theorem 7.10. The cyclotomic polynomial Ψn(X) is irreducible over Q.

Proof. Let f(X) be the minimal polynomial of ζn, a primitive nth root of unityover Q(X). We first note that by definition f(X) is monic, and thus sincef(X)|Xn − 1, we have

Xn − 1 = f(X)g(X) (7.2)

7.4. CYCLOTOMIC FIELDS 185

and f(X) and g(X) must be in Z[X].To prove that Ψn(x) is irreducible, we will actually prove that

Ψn(X) = f(X).

To prove the equality, it is enough to show that every root of Ψn(X) is a rootof f(X).

We need the following intermediate result: if p does not divide n, then

f(ζpn) = 0.

Let us prove this result. Suppose by contradiction that this is not the case,namely f(ζpn) 6= 0. By (7.2), we have

Xn − 1 = f(X)g(X),

which evaluated in X = ζpn yields

(ζpn)n − 1 = 0 = f(ζpn)g(ζ

pn)

implying by our assumption that f(ζpn) 6= 0 that

g(ζpn) = 0,

or in other words, ζn is a root of g(Xp). But by definition of minimal polynomial,we have that f(X) must then divide g(Xp), that is

g(Xp) = f(X)h(X), h(X) ∈ Z[X].

Since g(Xp), f(X) and h(X) are in Z[X], we can look at their reduction modulop, that is work in Fp[X]. We will denote p(X) the polynomial obtained fromp(X) by taking all its coefficients modulo p: if p(X) =

∑ni=0 aiX

i, then p(X) =∑n

i=0(ai mod p)Xi. Therefore

g(Xp) = f(X)h(X) ∈ Fp[X].

By working in Fp[X], we are now allowed to write that

g(Xp) = g(X)p

and thusg(X)p = f(X)h(X) ∈ Fp[X].

This tells us that any irreducible factor of f(X) divides g(X) and consequentlyf and g have a common factor. Looking at (7.2) in Fp[X] gives

Xn − 1 = f(X)h(X) ∈ Fp[X].

Since f and g have a common factor, Xn − 1 has a multiple root, which cannotbe since we have assumed that p does not divide n. This proves the claim.


To summarize, we have just proven that if p does not divide n, then f(ζpn) isanother root of f . Since all primitive nth roots of unity can be obtained fromζn by successive prime powers, we have that all primitive nth roots of unity areactually roots of f(X), and we know that there are ϕ(n) of them, which is alsothe degree of Ψn(X). This concludes the proof, since

deg f(X) ≥ ϕ(n) = deg(Ψn(X)) ⇒ f(X) = Ψn(X).

7.5 Solvability by radicals

The question of solvability by radicals is the one of solving polynomial equationsunder the restriction that we are only allowed to perform addition, subtraction,multiplication, division, and taking nth roots.

For example, we know (Fontana-Tartaglia, 1535) that for a cubic equation

X3 + pX = q,

the solution is given by

X =3

√

q

2+

√

p3

27+q2

4+

3

√

q

2−√

p3

27+q2

4.

By the 16th century all polynomial equations of degree smaller or equal to 4 weresolved. The natural question was then: what happens with quintic equations?Euler failed to give an answer, Lagrange (1770) proved that it depends on findingfunctions of the roots which are unchanged by certain permutations of the roots,and that this approach works up to degree 4 and fails for 5. Abel showed (1824)that quintics are insolvable by radicals. The next question thus became: decidewhether or not a given equation can be solved by radicals. Liouville (1843)found the answer in Galois’s papers.

The answer is to be found by connecting the problem with field theory asfollows. We first need to define the notion of a radical extension. Informally, aradical extension is obtained by adjoining a sequence of nth roots. For example,to get a radical extension of Q containing

3√11

5

√

7 +√3

2+

4

√

1 +3√4,

we must adjoin

α =3√11, β =

√3, γ =

5

√

7 + β

2, δ =

3√4, ǫ =

4√1 + δ.

This can be stated formally:

7.5. SOLVABILITY BY RADICALS 187

Definition 7.6. An extension E/F is radical if E = F (α1, . . . , αn) where forall i = 1, . . . , n, there exists an integer n(i) such that

αn(i)i ∈ F (α1, . . . , αi−1), i ≥ 2.

The αi’s are said to form a radical sequence for E/F .

Example 7.8. The expression

3√11

5

√

7 +√3

2+

4

√

1 +3√4

is contained in Q(α, β, γ, δ, ǫ), where

α3 = 11, β2 = 3, γ5 =7 + β

2, δ3 = 4, ǫ4 = 1 + δ.

Definition 7.7. Let f be a polynomial over a field F of characteristic zero (thisis a simplifying assumption). We say that f is solvable (soluble) by radicals ifthere exists a field E containing a splitting field for f such that E/F is a radicalextension.

We want to connect radical extensions and solvable groups. Here is the maintheorem:

Theorem 7.11. If F is a field of characteristic zero, and F ⊆ E ⊆ M whereM/F is a radical extension, then the Galois group of E/F is a solvable group.

Thus a solvable (by radicals) polynomial has a solvable Galois group (of asplitting field over the base field).

Recall that a group G is solvable if G has a normal series

{1} = Gr EGr−1 E . . .EG0 = G

with Gi/Gi+1 abelian. The proof takes some fair amount of work, though theidea is simple. A radical extension is a series of extensions by nth roots. Suchextensions have abelian Galois groups (to be proven though...), so the Galoisgroup of a radical extension is made up by fitting together a sequence of abeliangroups (unfortunately, the proof is not that simple...)

We can restate the above result in terms of polynomials.

Theorem 7.12. Let f be a polynomial over a field E of characteristic zero. Iff is solvable by radicals then its Galois group (that is the Galois group of itssplitting field) over E is a solvable group.

To find a polynomial which is not solvable by radicals, it suffices to find onewhose Galois group is not solvable.

Lemma 7.13. Let p be a prime, f an irreducible polynomial of degree p overQ. Suppose that f has precisely two non-real zeros in C. Then the Galois groupof f over Q is the symmetric group Sp.


Theorem 7.14. The polynomial X5−6X+3 over Q is not solvable by radicals.

The proof consists of showing that the polynomial is irreducible over Q, byEisenstein’s criterion. Then f has exactly three real zeros with multiplicity 1each, and the above lemma says that is Galois group is S5. To conclude, weneed to show that the symmetric group Sn is not solvable if n ≥ 5.

7.6 Solvability by ruler and compasses

The ancient Greek philosopher Plato believed that the only perfect figures werethe straight line and the circle, and this belief had a great impact in ancientGreek geometry: it restricted the instruments available for performing geomet-rical constructions to ruler and compasses.

Many constructions can be done just be using ruler and compasses, but threefamous constructions could not be performed:

• duplication of the cube: find a cube twice the volume of a given cube.

• trisection of the angle: find an angle 1/3 the size of a given angle.

• quadrature of the circle: find a square of area equal to those of a givencircle.

It is no wonder those problems remained unsolved (again, under these pla-tonic constraints) since we will see, using our modern tools, that none of themare possible.

We start by formalizing the intuitive idea of a ruler and compass construc-tion. Denote by P0 the set of points in R2.

• operation 1 (ruler): through any 2 points of P0, draw a straight line.

• operation 2 (compasses): draw a circle, whose center is a point of P0 andwhose radius is equal to the distance between some pairs of points in P0.

Definition 7.8. The points of intersection of any two distinct lines or circles,drawn using operations 1 and 2 are said to be constructible from P0 if thereexists a sequence r1, . . . , rn of points of R2 such that for each i = 1, . . . , n thepoint ri is constructible from the set P0 ∪ {r1, . . . , ri−1}, Pi = Pi−1 ∪ {ri}.

We can now bring field theory into play. With each stage, we associate thesubfield of R generated by the coordinates of the points constructed. Denote byK0 the subfield of R generated by the x- and y-coordinates of the points in P0.If ri has coordinates (xi, yi), then inductively we define

Ki = Ki−1(xi, yi)

to getK0 ⊆ K1 ⊆ . . . ⊆ Kn ⊆ R.

7.6. SOLVABILITY BY RULER AND COMPASSES 189

Lemma 7.15. With the above notation, xi and yi are zeros in Ki of quadraticpolynomials over Ki−1.

Proof. There are 3 cases to consider: line meets line, line meets circle and circlemeets circle. We only give the proof of line meets circle.

Take 3 points A = (p, q), B = (r, s), C = (t, u) in Ki−1, then draw a linebetween A and B, and a circle of center C with radius w. The equation of theline AB is

x− p

r − p=y − q

s− q

while the equation of the circle is

(x− t)2 + (y − u)2 = w2.

Solving them yields

(x− t)2 +

(s− q

r − p(x− p) + q − u

)2

= w2.

Now x, the first coordinate of the intersection point, is a zero of a quadraticpolynomial over Ki−1.

We note that fields obtained by adjoining the zeroes of a quadratic polyno-mial are extensions of degree 2.

Theorem 7.16. If r = (x, y) is constructible from a subset P0 ∈ R2, and if K0

is the subfield of R generated by the coordinates of the points of P0, then thedegrees [K0(x) : K0] and [K0(y) : K0] are powers of 2.

Proof. We have that

[Ki−1(xi) : Ki−1] = 1 or 2, [Ki−1(yi) : Ki−1] = 1 or 2.

Using multiplication of degrees, we get

[Ki−1(xi, yi) : Ki−1] = [Ki−1(xi, yi) : Ki−1(xi)][Ki−1(xi) : Ki−1] = 1 or 2 or 4

with Ki = Ki−1(xi, yi). Thus [Kn : K0] is a power of 2 implying that [Kn :K0(x)][K0(x) : K0] is a power of 2 from which we conclude that [K0(x) : K0] isa power of 2, and similarly for y.

We are now ready to discuss the impossibility proofs.

Theorem 7.17. The cube cannot be duplicated using ruler and compass con-structions.

Proof. Take a cube whose side is the unit interval, that is of volume 1. We haveP0 = {(0, 0), (1, 0)} and K0 = Q. If we could duplicate the cube, then we canconstruct a point (α, 0) such that the volume α3 is equal to 2, that is

α3 = 2.


Now [Q(α) : Q] is a power of 2, but α is a zero of t3 − 2 which is irreducible (byEisenstein) over Q. This gives that

[Q(α) : Q] = 3,

a contradiction to the fact that it should be a power of 2.

Theorem 7.18. The angle π/3 cannot be trisected using ruler and compassconstructions.

Proof. Constructing an angle trisecting π/3 is equal to constructing the point(α, 0) given (0, 0) and (1, 0) where α = cos(π/9). Knowing α = cos(π/9), wecan construct

β = 2 cos(π/9).

Using that cos(3θ) = 4 cos3(θ) − 3 cos(θ) and cos(3θ) = 1/2 when θ = π/9, wehave

1 = 8 cos3(θ)− 6 cos(θ) ⇒ β3 − 3β − 1 = 0.

Now f(t) = t3 − 3t− 1 is irreducible over Q (apply Eisenstein on f(t+ 1)) thus

[Q(β) : Q] = 3

contradicting the fact that it should be a power of 2.

Theorem 7.19. The circle cannot be squared using ruler and compass con-structions.

Proof. Without loss of generality, we assume that the circle is the unit circlecentered at (0, 0). Constructing a square with area π is equivalent to construct-ing a point (

√π, 0). Since the smallest field with 0 and 1 is Q, the field obtained

from adjoining (√π, 0) is Q(

√π). Thus [Q(

√π) : Q] should be a power of 2,

and in particular it should be algebraic, which is a contradiction (Lindeman’sTheorem shows the transcendence of π, 1882).

The main definitions and results of this chapter are

• (4.1). Definitions of: Galois extension, Galois group,fixed field.

• (4.2). The fundamental theorem of Galois theory,Galois groups of intermediate fields.

• (4.3). Characterization of finite fields, their Galoisgroup, their multiplicative group.

• (4.4). Definition of cyclotomic field, primitive rootof unity, cyclotomic polynomial. The Galois group ofa cyclotomic field.

Chapter 8Exercises on Galois Theory

Exercises marked by (*) are considered difficult.

8.1 Galois group and fixed fields

Exercise 96. Compute the Galois group of X4 − 2 over Q and F3, the finitefield with 3 elements.

Answer. Over Q, we have

X4 − 2 = (X2 −√2)(X2 +

√2) = (X − 21/4)(X + 21/4)(X − i21/4)(X + i21/4),

while over F3, let w be a root of the irreducible polynomial X2 + X + 2 = 0,then

w2 = −w + 1, w4 = −1, w8 = 1

and

X4 − 2 = X4 +1 = (X2 −w2)(X2 +w2) = (X −w)(X +w)(X −w3)(X +w3).

8.2 The fundamental Theorem of Galois theory

Exercise 97. 1. Compute the splitting field K of the polynomial f(x) =x4 − 2 ∈ Q(x).

2. Show that K is a Galois extension.

3. Compute the degree of K/Q.

4. Compute the Q-automorphisms of K.

5. Do you recognize Gal(K/Q)?

191

192 CHAPTER 8. EXERCISES ON GALOIS THEORY

6. What are all the subgroups of Gal(K/Q)?

7. What are all the intermediate subfields of K/Q?

8. Among the intermediate subfields, which are normal?Answer.

1. We have that f(x) = (x2−√2)(x2+

√2) = (x− 4

√2)(x+ 4

√2)(x+i 4

√2)(x−

i 4√2). Thus the splitting field of f is Q(i, 4

√2).

2. It is a splitting field thus K is normal, it is separable because Q is ofcharacteristic zero.

3. The degree is

[Q(4√2, i) : Q] = [Q(

4√2, i) : Q(

4√2)][Q(

4√2) : Q].

The minimum polynomial of i overQ( 4√2) is x2+1, so [Q( 4

√2, i) : Q( 4

√2)] =

2. Since f(x) is irreducible over Q (by Eisenstein), it is the minimal poly-nomial of 4

√2 over Q, thus [Q( 4

√2) : Q] = 4 and finally the total degree is

8.

4. There are 8 of them. We have

σ(i) = i, σ(4√2) = i

4√2,

and

τ(i) = −i, τ( 4√2) =

4√2

and we can find the others by combining these two, namely:

1 : 4√2 7→ 4

√2, i 7→ i

σ : 4√2 7→ i 4

√2 i 7→ i

σ2 : 4√2 7→ − 4

√2 i 7→ i

σ3 : 4√2 7→ −i 4

√2 i 7→ i

τ : 4√2 7→ 4

√2 i 7→ −i

στ : 4√2 7→ i 4

√2 i 7→ −i

σ2τ : 4√2 7→ − 4

√2 i 7→ −i

σ3τ : 4√2 7→ −i 4

√2 i 7→ −i

5. This is the dihedral group of order 8.

6. • order 8: G, order 1: {1}.• order 4: there are 3 of them

S = {1, σ, σ2, σ3} ≃ C4, T = {1, σ2, τ, σ2τ} ≃ C2×C2, U = {1, σ2, στ, σ3τ} ≃ C2×C2.


• order 2, there are 5 of them, all isomorphic to C2:

A = {1, σ2}, B = {1, τ}, C = {1, στ}, D = {1, σ2τ}, E = {1, σ3τ}.

7. By Galois correspondence, we obtain the intermediate fiels as fixed fieldsof the subgroups. The subfields of degree 2 are the easiest to find:

Q(i), Q(√2), Q(i

√2)

which are fixed by resp. S, T and U . By direct computation (that is,apply the automorphism on an element of the larger field, and solve theequation that describes that this element is fixed by this automorphism),we find that the others are:

Q((1 + i)4√2), Q(i,

√2), Q(

4√2).

fixed resp. by C, A and B.

8. The normal subgroups of G are G,S, T, U,A, I, thus their correspondingfixed fields are normal extensions of Q.

Exercise 98. Let K be the subfield of C generated over Q by i and√2.

1. Show that [K : Q] = 4.

2. Give a primitive element of K and its minimal polynomial.

3. Show that Gal(K/Q) ≃ (Z/2Z)2.

4. Give a list of all the subfields of K.

Answer.

1. Since K = Q(i,√2), we can first build Q(

√2)/Q which is of degree 2,

because x2 − 2 is irreducible, then we check that x2 +1 is irreducible overQ(

√2), so we obtain another extension of degree 2, by multiplicativity of

the degrees, this gives an extension of degree 4.

2. For example, ζ8, the primitive 8th root of unity, is a primitive element,with minimal polynomial x4 + 1.

3. The Galois group is given by {1, σ, τ, στ} where

σ : i 7→ −i,√2 7→

√2, τ : i 7→ i,

√2 7→ −

√2.

4. There is one for each subgroup of the Galois group. Since there are onlysubgroups of order 2 (but for the whole group and the trivial subgroup),we get 3 quadratic field extensions:

Q(i), Q(√2), Q(i

√2).


Exercise 99. 1. Show that X4 − 3 = 0 is irreducible over Q.

2. Compute the splitting field E of X4 − 3 = 0.

3. Compute the Galois group of E/Q.

4. Can you recognize this group?

5. Choose two proper, non-trivial subgroups of the Galois group above, andcompute their corresponding fixed subfields.

Answer.

1. Use Eisenstein with p = 3.

2. The roots of X4 − 3 are ij 4√3, j = 0, 1, 2, 3, thus the splitting field is

Q( 4√3, i).

3. As in previous exercise, with 4√3 instead of 4

√2.

4. It is the dihedral group.

5. Again as in previous exercise.

Exercise 100. Consider the field extensions M = Q(√2,√3) and E = M(α)

where α =√

(2 +√2)(3 +

√3).

1. Show that M is a Galois extension of Q with Galois group C2 × C2.

2. Denote by σ and τ the generators of the two cyclic groups of (1), so thatthe Galois group of M is written 〈τ〉 × 〈σ〉.

• Compute σ(α2)/α2 and deduce that α 6∈ M . What is the degree ofE over Q?

• Extend σ to an automorphism of E and show that this automorphismhas order 4.

• Similarly extend τ to an automorphism of E and compute its order.What is the Galois group of E over Q?

Answer.

1. M/Q is clearly Galois because it is separable (Q is of characteristic 0) andnormal.

2. • We have σ(α2)/α2 = (√2−1)2 thus σ(α2) = (α(

√2−1)2. If α were in

M , then σ(α) = ±α(√2−1) and σ2(α) = α(

√2−1)(−

√2−1) = −α,

a contradiction (σ2(α) = α).

• We have σ2(α) = −α thus σ4(α) = α, σ4|M = 1 and σ2 6= 1.


• We compute that τ(α) = 3−√3√

6α, we extend τ and its order is 4. We

then compute that

στ(α) =3−

√3

−√6

(√2− 1)α, τ(σ(α) = (

√2− 1)

3−√3√

6α

so σ and τ anticommute, and they are both of order 3, so the Galoisgroup is the quaternion group.

Exercise 101. Let L/K be a Galois extension of degree 8. We further assumethat there exists a subextensionM/K of degree 4 which is not a Galois extension.

• Show that the Galois group G of L/K cannot be abelian.

• Determine the Galois group G of L/K.

Answer.

• A subextension M/K of degree 4 which is not Galois, means that thereis a subgroup of order 2 which is not normal in G. Thus G cannot beabelian, since all subgroups of an abelian group are all normal.

• The only groups of order 8 which are not abelian are D4 and Q8. All thesubgroups of Q8 are normal, thus it must be D4.

Exercise 102. Assume that the polynomial X4+aX2+b ∈ Q[X] is irreducible.Prove that its Galois group is:

1. the Klein group if√b ∈ Q.

2. the cyclic group of order 4 if√a2 − 4b

√b ∈ Q.

Answer.

1. Set Y = X2, then

Y 2 + aY + b = (Y − y1)(Y − y2)

with

y1 =−a+

√a2 − 4b

2, y2 =

−a−√a2 − 4b

2

and X = ±√Y so that the four roots are ±√

y1,±√y2. Now y1y2 = b

thus √y1√y2 =

√b ∈ Q

and if σ(√y1) =

√y2, then we have that

σ(√y2) =

√b/σ(

√y1) =

√b/√y2 =

√y1

and all the elements of the Galois group have order 2, so that it must bethe Klein group.


2. We havey1 − y2 =

√

a2 − 4b,

thus √y1√y2(y1 − y2) =

√b√

a2 − 4b ∈ Q.

Now take σ(√y1) =

√y2 and if it were of order 2, then σ(

√y2) =

√y1 and

σ(√y1√y2(y1 − y2)) =

√y1√y2(y2 − y1)

which contradicts that√y1√y2(y1 − y2) ∈ Q thus σ is of order 4 and the

Galois group must be the cyclic group of order 4.

8.3 Finite fields

Exercise 103. Identify the finite fields Z[i]/(2 + i) and Z[i]/(7).

Answer. F5 and F49

Exercise 104. Consider the following two polynomials p(x) = x2−x−1 ∈ F3[x]and q(x) = x2 + 1 ∈ F3[x]. Consider the fields F3[x]/(p(x)) ≃ F3(α) wherep(α) = 0 and F3[x]/(q(x)) ≃ F3(β) where q(β) = 0.

1. Compute (α+ 1)2.

2. Deduce that the two fields F3(α) and F3(β) are isomorphic.

Answer.

1. We have (α+ 1)2 = α2 − α+ 1 = (α+ 1)− α+ 1 = 2 = −1.

2. We have that β2 = −1 by definition of β and we have just shown abovethat (α+1)2 = −1, thus it is natural to map β to α+1, that is f : F3(β) →F3(α), a + bβ 7→ a + b(α + 1). Check that f is a ring homomorphism.Then argue that a field homomorphism is always injective, and that bothfields have same number of elements.

Exercise 105. Let F2 be the finite field with two elements.

1. Show that F2(β) = F2[X]/(q(X)) is a finite field, where q(X) = X2+X+1and q(β) = 0.

2. Consider the polynomial r(Y ) = Y 2 + Y + β ∈ F2(β)[Y ], and set L =F2(β)[Y ]/(r(Y )).

• Is L a field? Justify your answer.

• What is the cardinality of L? What is its characteristic? Justify youranswers.

Answer.


1. It is enough to show that q(X) is irreducible over F2, this generates thefinite field F4 ≃ F2(β).

2. • We have to see if r(Y ) is irreducible over F4. It is enough to evaluateit in β and β + 1 to see that it is not zero.

• This creates an extension of degree 2 of F4, that is 16 elements. Ithas characteristic 2.

Exercise 106. • Let Fp be a finite field, p ≥ 3 a prime number. Show thatthe sum of all the elements of Fp is 0.

• Let q = pn, p a prime. Show that if q 6= 2, then the sum of all elements ofFq is 0.

• Let q = pn, p a prime. Show that the product of all the non-zero elementsof a finite field Fq is -1.

Answer.

• There are many ways of doing that. Modulo p, one could simply noticethat 1+2+ . . .+p−1 is p(p−1)/2, if p ≥ 3, p is an odd prime, thus p−1is even, (p− 1)/2 is an integer and thus mod p we do get 0.

• An element a in Fq satisfies that apn

= a, that is, it is a root of Xpn −X.Now all the roots of this polynomial exactly coincide with the elements ofFq, that is, we can write

Xpn −X =∏

a∈Fq

(X − a).

If we develop the product, we get that the term inXpn−1 has as coefficientsexactly the sum of the elements of Fq, which is thus 0.

• This follows from above. Now we just factor X from the polynomialXpn −X to get

Xpn−1 − 1 =∏

a∈F∗

q

(X − a).

Now −1 corresponds to the constant term of the product, which is exactlythe product over all non-zero elements of the finite field.

Exercise 107. Consider the finite fields F2,F3 and F4, and the polynomialP (Y ) = Y 3 + Y + 1. Over which of these finite fields is P (Y ) irreducible? Ifpossible, construct the corresponding field extension.

Answer. Since this polynomial is of degree 3, if it is reducible, that meansthere is at least one linear term, that is one root in the base field. It is thusirreducible over F2, however over F3, we have that P (1) = 0, and over F4, wehave no root. Over F2, we get an extension of degree 3, that is F8, over F4, weget an extension of degree 3, that is F43 .


4.4 Cyclotomic fields

Exercise 108. Let ζ be a primitive 20th root of unity in C, and let E = Q(ζ).

• Compute the Galois group Gal(E/Q).

• How many subfields of E are there which are quadratic extensions of Q?

• Determine the irreducible polynomial of ζ over Q.

Answer.

• We know that Gal(E/Q) ≃ (Z/20Z)∗.

• There are 3 of them: Q(i√5), Q(

√5) and Q(i).

• It is X8 −X6 +X4 −X2 + 1.

4.5 Solvability by radicals

4.6 Solvability by ruler and compasses

Exercise 109. True/False.

Q1. An extension having Galois group of order 1 is normal.

Answer.

Q1. It’s false! If there is only one element, then it’s the identity. Again Q(α)with α3 = 2 has a Galois group with only the identity, and it is not normal!

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